Question

Two iron solid discs of negligible thickness have radii \( R_1 \) and \( R_2 \) and moment of inertia \( I_1 \) and \( I_2 \), respectively. For \( R_2 = 2R_1 \), the ratio of \( I_1 \) and \( I_2 \) would be \( \frac{1}{x} \), where \( x \) is:

Hint:

Remember that the moment of inertia scales with the square of the radius for simple geometric bodies like discs and spheres.

Answer 1

Correct Answer: 16

Step 1: Formula for Moment of Inertia of a Solid Disc

The moment of inertia of a solid disc about its central axis (perpendicular to its plane) is: \[ I = \frac{1}{2}MR^2 \]

Step 2: Moment of Inertia of Discs \( I_1 \) and \( I_2 \)

For the first disc: \[ I_1 = \frac{1}{2} M_1 R_1^2 \] For the second disc: \[ I_2 = \frac{1}{2} M_2 R_2^2 \]

Step 3: Given Condition

It is given that: \[ R_2 = 2R_1 \]

Since the discs are made of iron (same material), the mass is proportional to the area (assuming thickness is negligible). So: \[ \frac{M_2}{M_1} = \left( \frac{R_2}{R_1} \right)^2 = \left( \frac{2R_1}{R_1} \right)^2 = 4 \] \[ M_2 = 4M_1 \]

Step 4: Finding the Ratio of \( I_1 \) and \( I_2 \)

Substituting the values: \[ I_1 = \frac{1}{2} M_1 R_1^2 \] \[ I_2 = \frac{1}{2} \times 4M_1 \times (2R_1)^2 = \frac{1}{2} \times 4M_1 \times 4R_1^2 = 8M_1R_1^2 \]

Ratio \( \frac{I_1}{I_2} \): \[ \frac{I_1}{I_2} = \frac{\frac{1}{2} M_1 R_1^2}{8M_1R_1^2} = \frac{1}{2 \times 8} = \frac{1}{16} \]

Therefore, \[ \boldsymbol{x = 16} \]

Answer 2

Step 1: Understand the question.
We are given two solid iron discs with radii \( R_1 \) and \( R_2 \), and corresponding moments of inertia \( I_1 \) and \( I_2 \). Both have negligible thickness, meaning they can be treated as uniform solid discs. It is given that \( R_2 = 2R_1 \), and we need to find the ratio of \( I_1 \) to \( I_2 \), which is \( \frac{1}{x} \).

Step 2: Recall the moment of inertia of a solid disc.
The moment of inertia of a solid disc about its central axis (perpendicular to the plane of the disc) is given by:
\[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass of the disc and \( R \) is its radius.

Step 3: Express the mass in terms of density and radius.
Since both discs are made of the same material and have negligible thickness, the mass is proportional to the area (since thickness and density are constant):
\[ M \propto R^2 \] Hence, we can write:
\[ I \propto M R^2 \propto R^2 \times R^2 = R^4 \] So, the moment of inertia of a solid disc varies as the fourth power of its radius.

Step 4: Find the ratio of \( I_1 \) and \( I_2 \).
\[ \frac{I_1}{I_2} = \left( \frac{R_1}{R_2} \right)^4 \] Substitute \( R_2 = 2R_1 \):
\[ \frac{I_1}{I_2} = \left( \frac{R_1}{2R_1} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16} \] Hence, the ratio is \( \frac{1}{x} = \frac{1}{16} \).

Step 5: Conclusion.
\[ x = 16 \]

Final Answer:
\[ \boxed{x = 16} \]